o is any point inside a rectangle ABCD. Prove that OB^2+OD^2=OA^2+OC^2

1.	o is any point inside a rectangle ABCD. Prove that OB^2+OD^2=OA^2+OC^2
Answer» Let ABCD be the given rectangle and let O be a point within it. Join OA, OB, OC and OD.Trough O, draw {tex}E O F \\\| A B.{/tex}\xa0Then, ABFE is a rectangle.In right triangles {tex}\\Delta{/tex}OEA and {tex}\\Delta{/tex}OFC, we have\xa0OA2 = OE2+ AE2 and OC2 = OF2\xa0+ CF2{tex}\\Rightarrow{/tex}\xa0OA2 + OC2 = (OE2 + AE2) +(OF2+ CF2){tex}\\Rightarrow{/tex}\xa0OA2 + OC2\xa0= OE2\xa0+ OF2\xa0+ AE2 + CF2\xa0...... (1)Now, in right triangles {tex}\\Delta{/tex}\xa0OFB and {tex}\\Delta{/tex}\xa0ODE, we haveOB2= OF2+FB2 and OD2= OE2+DE2{tex}\\Rightarrow{/tex}\xa0OB2 + OD2 = (OF2 + FB2) + (OE2+ DE2){tex}\\Rightarrow{/tex}\xa0OB2+ OD2= OE2+ OF2+DE2+BF2{tex}\\Rightarrow{/tex}\xa0OB2+ OD2\xa0= OE2+ OF2+ CF2+AE2 [{tex}\\because{/tex}\xa0DE= CF and AE = B F] ......(2)From (i) and (ii), we getOA2 + OC2 = OB2 + OD2

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